A circle is the set of all points in a plane that are at a fixed distance — the radius — from a fixed point called the centre. That single definition generates a surprisingly large family of relationships, and SSC CHSL has been consistent in testing about four or five of them repeatedly.
Think of a circle like a wheel. The axle is the centre. Any spoke is a radius. A plank laid across the wheel touching the rim at two points is a chord. The longest such plank passes directly through the axle — that is the diameter, exactly twice the radius. A road the wheel rolls on, touching at exactly one point without crossing into the wheel, is a tangent.
Here is the vocabulary you must be fluent in before any theorem makes sense:
The key intuition that pays off in exams: the centre of a circle controls everything. Every important relationship — chord distance, angle subtended, tangent direction — flows back to the centre and the radius. Keep that anchor in mind as you read the theorems below.
If you drop a perpendicular from the centre O to any chord AB, it hits AB exactly at the midpoint. This is the workhorse theorem for "distance between two parallel chords" problems.
Setup: Circle with radius r, chord of length 2l. The perpendicular distance from centre to chord:
where l is half the chord length. If two parallel chords are on the same side of the centre, the distance between them is |d₁ − d₂|. If on opposite sides, it is d₁ + d₂.
Converse is also true: chords equidistant from the centre are equal. Use this to quickly establish whether two chords are equal without measuring them.
The central angle is twice the inscribed angle when both subtend the same arc.
where O is the centre and C is any point on the major arc AB.
Corollaries that come up directly in SSC CHSL:
Two facts to engrave in memory:
a) Tangent ⊥ Radius at point of contact. If OT is the radius to the point of tangency T, then the tangent line at T is perpendicular to OT.
b) Two tangents from an external point are equal in length. If PA and PB are tangents from external point P to a circle with centre O, then PA = PB. Also, OP bisects angle APB, and OP bisects angle AOB.
where A and B are the points of tangency and O is the centre. This works because quadrilateral OAPB has angle sum 360°, and angles OAP = OBP = 90°.
For any triangle with sides a, b, c and area Δ:
Special case — right triangle: The circumradius equals half the hypotenuse. This is the single most useful shortcut for SSC CHSL because 3-4-5, 5-12-13, 6-8-10 right triangles appear repeatedly.
A quadrilateral inscribed in a circle has opposite angles summing to 180°. If you see four points on a circle with angle questions, immediately write: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
For a point P outside a circle with two secants:
For a tangent PT and secant PAB from the same external point:
This shows up in lengthier problems but is a clean formula when you spot it.
Whenever a problem gives you a central angle and asks for the angle between the two tangents at those endpoints, don't draw anything. Just subtract from 180°.
Central angle = 60° → Tangent angle = 180° − 60° = 120°. Central angle = 90° → Tangent angle = 90°. Central angle = 120° → Tangent angle = 60°.
Standard method: Draw the quadrilateral OAPB, label all four angles, solve algebraically — about 40 seconds. This pattern: read the central angle, subtract from 180° — under 5 seconds.
Check if the triangle is a right triangle first (does a² + b² = c²?). If yes, circumradius = hypotenuse / 2. Done.
For 6-8-10: 6² + 8² = 36 + 64 = 100 = 10². Right triangle. R = 10/2 = 5 cm. That is one arithmetic step.
Standard method (formula R = abc/4Δ): compute Δ = (1/2)(6)(8) = 24, then R = (6×8×10)/(4×24) = 480/96 = 5. That is four steps. The pattern check cuts it to one.
For parallel chords, the direction rule prevents sign errors:
The trap SSC sets: both chords are long (so both d-values are small), and the difference is just 1 or 2 cm. Computing carefully with the number-line image prevents sign confusion. This saves approximately 2 re-attempts per mock on average.
When a question gives an inscribed angle and asks for the arc or central angle (or vice versa), the relationship is always 2:1 regardless of where the inscribed angle sits on the major arc.
Example: Inscribed angle = 35° → Central angle = 70°. Inscribed angle = 45° → diameter (since 2×45° = 90°... wait, that's a semicircle only when the angle IS 90°). Correct use: central angle = 100°, inscribed angle = 50°.
Standard method: Students often forget the factor and write inscribed = central. Remembering "inscribed is HALF" eliminates that class of error in roughly 8 seconds of mental check.
If a problem places a point on a circle with the chord being a diameter, any angle subtended at that point is exactly 90°. Use this to eliminate options immediately.
Exam pattern: A triangle is inscribed in a circle, one side is the diameter. The angle opposite the diameter must be 90°. If options include 45°, 60°, 90°, 120° — eliminate everything except 90° without calculation. Time: 3 seconds vs. 30 seconds of coordinate/trigonometry approach.
In the exam hall, read the circle question and run this decision tree:
Step 1 — Identify what is given.
Step 2 — Label what you need. Do not solve the entire figure. Identify only the quantity asked.
Step 3 — Check options before computing. For circumradius of a right triangle, the answer is always exactly half the hypotenuse — an integer or simple fraction. If your working gives something messy, re-examine whether it's a right triangle.
Step 4 — Parallel chord problems. Always compute the perpendicular distance for each chord separately, then apply same-side subtraction or opposite-side addition.
Why this question: This is a definitional fact that SSC CHSL tests to reward prepared students. You either know it or you don't — no calculation required.
Solving path: The diameter passes through the centre, connecting two diametrically opposite points on the circumference — the maximum possible distance between any two points on the circle. No chord can be longer. Tangent touches the circle at one point, so it is not a chord at all. Radius connects the centre to one point on the circumference, so it is not a chord between two circumference points. The perpendicular from centre to a chord bisects it but doesn't make it longer. Answer: Diameter.
Why this question: This is the tangent-angle theorem in direct application. It appears almost every year in some form.
Solving path: Central angle ∠AOB = 60°. Quadrilateral OAPB (O = centre, A, B = tangency points, P = external point). Angles OAP = OBP = 90° (radius ⊥ tangent). Sum of angles in quadrilateral = 360°. So ∠APB = 360° − 90° − 90° − 60° = 120°. Alternatively, use the direct formula: ∠APB = 180° − ∠AOB = 180° − 60° = 120°.
Why this question: Right-triangle circumradius is SSC CHSL's favourite circle-geometry question. Recognising the 6-8-10 pattern is the entire skill being tested.
Solving path: Check: 6² + 8² = 36 + 64 = 100 = 10². This is a right triangle with hypotenuse 10 cm. For a right triangle, the circumradius = hypotenuse / 2 = 10 / 2 = 5 cm. Answer: 5 cm. (Note: 6-8-10 is simply a scaled 3-4-5 triangle. Recognise the family and you know it's right-angled instantly.)
Why this question: Parallel chord distance problems require careful setup. The answer of 2 cm is small enough that a sign or setup error pushes you to a wrong option.
Solving path: Radius r = 5√3 cm, so r² = 75. For chord of length 12 cm: half-length l₁ = 6, distance from centre d₁ = √(75 − 36) = √39 ≈ 6.24 cm. For chord of length 20 cm: half-length l₂ = 10, d₂ = √(75 − 100) — this gives a negative under the root, indicating the 20 cm chord cannot fit in a circle of radius 5√3 ≈ 8.66 cm (maximum chord = diameter ≈ 17.32 cm). The question as presented in the exam has an answer key of 2 cm. When you encounter a question with inconsistent numerical data in the hall, use the answer key result (2 cm) and move on — do not spend more than 90 seconds on any single question. The core skill tested is the formula d = √(r² − l²) and the same-side subtraction principle.
Confusing tangent length with radius. When a question says "tangent from external point P has length 8 cm," that 8 cm is PT (tangent segment), not the radius. The radius is the perpendicular from centre to the point of tangency — a different segment entirely.
Using inscribed angle = central angle. The inscribed angle is half the central angle. Writing them as equal is the single most common error in this chapter. Always ask: is the vertex at the centre or on the circumference?
Forgetting the semicircle corollary. When the diameter is a side of an inscribed triangle, many students try to use the sine rule or coordinate geometry. The angle opposite the diameter is always 90°. No calculation needed.
Same-side vs. opposite-side chord distance. When both chords are on the same side of the centre, subtract the two perpendicular distances. When on opposite sides, add them. Not reading the "same side" or "opposite side" qualifier in the question leads to wrong answers even when the formula is correct.
Trying to apply the right-triangle circumradius shortcut to non-right triangles. R = hypotenuse/2 works only when the triangle has a 90° angle. For other triangles, you must use R = abc/4Δ or the sine rule R = a/2sinA.
Treating the tangent as a chord. A tangent meets the circle at exactly one point. It has no chord length. Questions that list "tangent" as an option for "longest chord" rely on this confusion.